First Order Non-Linear Differential Equation

**Belowzero78** · October 3rd 2010, 10:27 AM

Question: Solve the Initial value problem:

$(5t^4y^{-3}+7t^6y^7)dt + (-3t^5y^{-4} + 7t^7y^6)dy = 0, y(1) = 1$ .

Express your answer in the form $F(t,y) = 2$ , where $F(t,y)$ has no constant term.

$F(t,y)=$ __________________ $= 2$

Attempt at the question:

First of all I checked the terms to see if they were exact and they are not. I took the integral with respect to $t$ for $M(t,y)$ which is the first part of the equation and it came out to be $t^5y^{-3} + t^7y^7 + k(y)$ . Afterwards I took the partial with respect to y of that integral we just got which was: $-3t^5y^{-4} + 7t^6y^7 + k'(y)$ . I now would equate both these equations together and would cancel terms. The result gives me $k'(y) = 0$ . So, my equation $F(t,y) = t^5y^{-3} + t^7y^7$ . But my question now is, how do I solve using the initial values since my equation has both terms $t$ and $y$ ?

Thanks for everyone who contributes!

**TheEmptySet** · October 3rd 2010, 10:44 AM

This is an implicitly defined curve. You are told that the solution should look like

$F(y,t)=2$ this must hold for all y and t

You have $F(y,t)=\frac{t^5}{y^3}+t^7y^7=2=F(1,1)$ so the implict solution is

$\frac{t^5}{y^3}+t^7y^7=2$ we can check this by taking the derivative

$5t^4y^{-3}-(3t^5y^{-4})\frac{dy}{dt}+7t^6y^7+(7t^7y^6)\frac{dy}{dt}=0$

$(-3t^5y^{-4}+7t^7y^6)dy+(5t^4y^{-3}+7t^6y^7)dt=0$

**Belowzero78** · October 3rd 2010, 10:46 AM

Ok, so my answer that i reached is correct. Essentially, what does no constant term mean? I just want to clarify that for my understanding!

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